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A soap film circle in $x-y$ plane with center at origin can be carefully pricked with a blunt soapy pin at center and drawn out a little bit on $z$-axis forming a surface of revolution somewhat like a tent roof. What shape/equation does it have before onset of Goldschmidt instability collapse? It is easy to practically check out its formation with liquid detergent.

The $y = c \cosh (x/c)$ classic catenoid minimal area case between two rings with $c$ depending on surface tension etc. is the only symmetric case mentioned in text books.

Narasimham
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    I must say I'm surprised at the vote to close. This looks like a very interesting question to me. – TonyK Aug 11 '14 at 16:55
  • Pics or it didn't happen. Seriously, catenoid is the only rotationally symmetric minimal surface, apart from the plane. I think this is the shape you are seeing. It's just that the catenoid is very thin-waisted, the waist being the pin that you lift it with. An infinitesimally thin pin would not lift the soap film at all. Of course, the shape will be only the part of catenoid from waist down. –  Aug 11 '14 at 17:06
  • Goldschmidt limit for film breakup is x/c ~ 1.2 (by solving y/x = dy/dx). What is practically seen is much more in proportion. – Narasimham Aug 11 '14 at 19:11

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I think 900-sillyname has it right in the comments. If the pin were not blunt, the minimal-area surface would consist of a flat circle connected to the pin by a cylinder of radius zero, wouldn't it? And this is not physically realisable. So the experimental surface that you get is really a catenoid, with a non-zero radius at the apparent apex.

And the sharper the pin, the flatter the surface becomes (and the thinner the 'stem'), until the surface tension in the stem causes the film to break. I would expect this to happen well before surface started to look really flat.

TonyK
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